A particle starts from a point O and moves in a straight line. The velocity vms¯¹ of the particle at time ts after leaving O is given by v = k(3t² – 2t³), where k is a constant.

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

1. [*M1] For attempt at integration (The power of t must increase by 1 with a change of coefficient in at least 1 term. Allow if k is omitted.)
2. [A1] k(t³ - 1/2t⁴) [+C] (Allow unsimplified)
3. [DM1] k(8 - 1/2 * 16) [-0] (For use of limits 0 and 2 or substituting t = 2 OR equate to 0, then solve a quartic equation in t. Allow if k is omitted.)
4. [A1] 0 (AG, CWO including stating C = 0 if not using limits)
5. [M1] v = 0 when k(3t² - 2t³) = 0 (For solving for t)
6. [A1] Leading to t = 1.5 [or t = 0] (A0 for other solutions not discarded)
7. [M1] a = k(6t - 6t²) (For differentiation, the power of t must decrease by 1 with a change of coefficient in at least 1 term. Allow if k is omitted)
8. [A1] [-13.5=k(6*1.5-6*1.5²) => -13.5=k(9-13.5)=> ] k = 3
9. [DM1] Distance from t = 0 to t = 1.5 is 3 ∫(t³ - 1/2t⁴) with limits 0 and 1.5 (For use of limits. 2*integral from 0 to 1.5 of (t³ - 1/2t⁴) or 2 * (1.5³ - 1/2 * 1.5⁴) or both integrals. Dependent on first M in part (a), unless they restart)
10. [A1] [= 2.53125] So total distance = 2 * 2.53125 = 5.06 m (Allow distance = 81/16 = 5 1/16. If DM0 then SCB1 for 5.06 without working)